Grade 11 Mathematics worksheet on Equations and Inequalities with sample questions and solutions.

Problem Analysis:

The image contains a worksheet on solving equations and inequalities, focusing on quadratic equations. The tasks are divided into several sections:

1. Solving quadratic equations using factorization or simplification.
2. Solving fractional equations using factorization or simplification.
3. Solving quadratic equations by completing the square.
4. Solving equations by squaring both sides.

Since the image is quite extensive, I will solve one example from each section to demonstrate the methods. If you need solutions for specific questions, please let me know!

---

Section 1: Solving Quadratic Equations Using Factorization or Simplification

#### Example: Question 1.1
Solve for \( x \):
\[ 3x^2 - 27 = 0 \]

Solution:
1. Start with the given equation:
\[
3x^2 - 27 = 0
\]
2. Factor out the common term (3):
\[
3(x^2 - 9) = 0
\]
3. Recognize that \( x^2 - 9 \) is a difference of squares:
\[
x^2 - 9 = (x - 3)(x + 3)
\]
4. Substitute back:
\[
3(x - 3)(x + 3) = 0
\]
5. Set each factor equal to zero:
\[
x - 3 = 0 \quad \text{or} \quad x + 3 = 0
\]
6. Solve for \( x \):
\[
x = 3 \quad \text{or} \quad x = -3
\]

Answer:
\[
\boxed{x = 3 \text{ or } x = -3}
\]

---

Section 2: Solving Fractional Equations Using Factorization or Simplification

#### Example: Question 2.1
Solve for \( x \):
\[
\frac{10}{3(x-1)} + \frac{1}{3} = \frac{4}{x-2}
\]

Solution:
1. Start with the given equation:
\[
\frac{10}{3(x-1)} + \frac{1}{3} = \frac{4}{x-2}
\]
2. Find a common denominator for the left-hand side. The denominators are \( 3(x-1) \) and \( 3 \), so the common denominator is \( 3(x-1) \):
\[
\frac{10}{3(x-1)} + \frac{1}{3} = \frac{10}{3(x-1)} + \frac{(x-1)}{3(x-1)} = \frac{10 + (x-1)}{3(x-1)} = \frac{x + 9}{3(x-1)}
\]
3. Rewrite the equation:
\[
\frac{x + 9}{3(x-1)} = \frac{4}{x-2}
\]
4. Cross-multiply to eliminate the fractions:
\[
(x + 9)(x - 2) = 4 \cdot 3(x - 1)
\]
5. Expand both sides:
\[
x^2 + 9x - 2x - 18 = 12(x - 1)
\]
\[
x^2 + 7x - 18 = 12x - 12
\]
6. Move all terms to one side to form a quadratic equation:
\[
x^2 + 7x - 18 - 12x + 12 = 0
\]
\[
x^2 - 5x - 6 = 0
\]
7. Factor the quadratic equation:
\[
x^2 - 5x - 6 = (x - 6)(x + 1) = 0
\]
8. Set each factor equal to zero:
\[
x - 6 = 0 \quad \text{or} \quad x + 1 = 0
\]
\[
x = 6 \quad \text{or} \quad x = -1
\]

Answer:
\[
\boxed{x = 6 \text{ or } x = -1}
\]

---

Section 3: Solving Quadratic Equations by Completing the Square

#### Example: Question 3.1
Solve for \( x \) by completing the square:
\[
x^2 + 2x = 24
\]

Solution:
1. Start with the given equation:
\[
x^2 + 2x = 24
\]
2. To complete the square, take the coefficient of \( x \) (which is 2), divide it by 2, and square it:
\[
\left(\frac{2}{2}\right)^2 = 1
\]
3. Add this value to both sides of the equation:
\[
x^2 + 2x + 1 = 24 + 1
\]
\[
x^2 + 2x + 1 = 25
\]
4. The left-hand side is now a perfect square:
\[
(x + 1)^2 = 25
\]
5. Take the square root of both sides:
\[
x + 1 = \pm \sqrt{25}
\]
\[
x + 1 = \pm 5
\]
6. Solve for \( x \):
\[
x + 1 = 5 \quad \text{or} \quad x + 1 = -5
\]
\[
x = 4 \quad \text{or} \quad x = -6
\]

Answer:
\[
\boxed{x = 4 \text{ or } x = -6}
\]

---

Section 4: Solving Equations by Squaring Both Sides

#### Example: Question 4.1
Solve for \( x \) by squaring both sides:
\[
\sqrt{5 - x} = x + 1
\]

Solution:
1. Start with the given equation:
\[
\sqrt{5 - x} = x + 1
\]
2. Square both sides to eliminate the square root:
\[
(\sqrt{5 - x})^2 = (x + 1)^2
\]
\[
5 - x = x^2 + 2x + 1
\]
3. Move all terms to one side to form a quadratic equation:
\[
5 - x - x^2 - 2x - 1 = 0
\]
\[
-x^2 - 3x + 4 = 0
\]
4. Multiply through by \(-1\) to simplify:
\[
x^2 + 3x - 4 = 0
\]
5. Factor the quadratic equation:
\[
x^2 + 3x - 4 = (x + 4)(x - 1) = 0
\]
6. Set each factor equal to zero:
\[
x + 4 = 0 \quad \text{or} \quad x - 1 = 0
\]
\[
x = -4 \quad \text{or} \quad x = 1
\]
7. Check for extraneous solutions by substituting back into the original equation:
- For \( x = -4 \):
\[
\sqrt{5 - (-4)} = -4 + 1 \implies \sqrt{9} = -3 \quad \text{(not valid)}
\]
- For \( x = 1 \):
\[
\sqrt{5 - 1} = 1 + 1 \implies \sqrt{4} = 2 \quad \text{(valid)}
\]

Answer:
\[
\boxed{x = 1}
\]

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Final Answers:

1. Question 1.1: \(\boxed{x = 3 \text{ or } x = -3}\)
2. Question 2.1: \(\boxed{x = 6 \text{ or } x = -1}\)
3. Question 3.1: \(\boxed{x = 4 \text{ or } x = -6}\)
4. Question 4.1: \(\boxed{x = 1}\)

If you need solutions for other specific questions, feel free to ask!